sin2xsinx=cosx?

Question

Help? I know its a double angle equation, but none of the formulas seem to get the correct answer. The only instructions are to find the exact solutions of the equation in the interval [0,2pi)

Answer 1

(sin 2x)(sin x) = cos x
(2sin x cos x)(sin x) = cos x
(2sin² x)(cos x) - cos x = 0
(2sin² x - 1)(cos x) = 0
if cos x = 0, x = π/2, 3π/2
if 2sin² x - 1 = 0
sin² x = 1/2
sin x = ±√2 / 2
x = π/4, 3π/4, 5π/4, 7π/4

Answer 2

Observe that sin2xsinx = cosx can be re-written as:
(2sinxcosx)sinx = cosx (used dbl angle id for 1st factor)
2sin^2xcos = cosx
2sin^2xcosx - cosx = 0
cosx(2sin^2x -1) = 0.
Hence, either cosx = 0 (Case 1) or 2sin^2x -1 = 0 (Case 2) by the zero factor law.

For Case 1: cosx = 0 and x in [0,2pi) has roots x = pi/2 and x = 3pi/2.

For Case 2: 2sin^2x -1 = 0 ==> sin^2x = 1/2
==> sinx = +- 1/sqrt(2)
==> sinx = +- sqrt(2)/2
==> sinx = sqrt(2)/2 or sinx = -sqrt(2)/2
==> x = pi/4, 3pi/4 or x= 5pi/4 or 7pi/4 since x in [0,2pi).

The given equation therefore has six (6) roots in [0,2pi). These are x = pi/2, 3pi/2, pi/4, 3pi/4, 5pi/4, and 7pi/4.

Verify:
For x= pi/2:
sin2(pi/2)sin(pi/2) = sin(pi)sin(pi/2) = 0(1)
= 0 = cos(pi/2)

For x=3pi/2:
sin2(3pi/2)sin(3pi/2) = sin(3pi)sin(3pi/2) = (0)(-1)
= 0 = cos(3pi/2)

For x=pi/4:
sin2(pi/4)sin(pi/4) = sin(pi/2)sin(pi/4) = (1)(sqrt(2)/2)
= sqrt(2)/2 = cos(pi/4)

For x=3pi/4:
sin2(3pi/4)sin(3pi/4) = sin(3pi/2)sin(3pi/4)
= (-1)(sqrt(2)/2) = -sqrt(2)/2 = cos(3pi/4)

Similarly for x = 5pi/4 and x = 7pi/4.

Answer 3

sin 2x = 2 sin x cos x.
So your equation becomes
2 sin x cos x sin x = cos x.
Now we have to check 2 possibilities:
1) cos x = 0. This gives
x = π/2 or 3π/2.

2) cos x <> 0.
Now
2 sin² x = 1
sin² x = 1/2
sin x = 1/√2 or sin x = -1/√2.
This happens for x = π/4, 3π/4, 5π/4 and 7π/4.

Answer 4

sin(2x)sinx = cosx
2sinxcosxsinx = cosx ( Divide both sides by cosx)
2sin²x = 1
sin²x = ½
sinx = √½
sinx=±√2 / 2

x = 45, 135 , 225 , 315 Degrees

x=π/4 , 3π/4 , 5π/4 , 7π/4 Rad

Answer 5

Hey,
You can use the identity sin(2x) = 2*sin(x)*cos(x)
take out the cos(x) in both sides of the equation.
you have 2*[sin(x)^2) = 1

which means that sin(x) = + / - 1 / sqrt(2)

the exact solutions (in degrees) are 45, 135, 225, 315

Answer 6

sin2xsinx = cosx
(2sinxcosx)sinx = cosx

divide both sides by cosx

2sinxsinx = 1
sinxsinx = 1/2
sinx = sqrt(1/2)


One of the answers is 45 degrees. There is another one
as well, left for you to figure out.